4t^2+10t-100=0

Solution for 4t^2+10t-100=0 equation:

4t^2+10t-100=0

a = 4; b = 10; c = -100;
Δ = b2-4ac
Δ = 102-4·4·(-100)
Δ = 1700
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1700}=\sqrt{100*17}=\sqrt{100}*\sqrt{17}=10\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{17}}{2*4}=\frac{-10-10\sqrt{17}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{17}}{2*4}=\frac{-10+10\sqrt{17}}{8}
$